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HELPSubstitution and Integral by Parts

  1. Dec 1, 2009 #1
    I'm having big trouble when trying to figure this integral out. Please help!

    Integral (from 0 to infinity): ((x^2)/a)*e^[(-x^2)/2a] dx a is a constant


    Thanks in advance!!
     
    Last edited: Dec 1, 2009
  2. jcsd
  3. Dec 1, 2009 #2

    Mark44

    Staff: Mentor

    I would start with u = x and dv = (1/a)x*e-x2/(2a)dx.
     
  4. Dec 1, 2009 #3
    OHHHH! Thank you sooo much!! I'v been thinking my head off by trying to do substitution!!

    And I have one more question, I'm required to calculate E(x2) too, so now instead of (X^2)/a for the first part, it becomes (x^3)/a, and the 2nd part remains the same. How do I approach this one?

    Thank you very much!
     
  5. Dec 1, 2009 #4
    Oh wait a minute, after doing the parts, since V = - (e-x2/(2a)), now applying the fomula, uv - integral v du, how do I solve Integral - (e-x2/(2a)) dx ?
     
  6. Dec 1, 2009 #5
    Hi Mark44, I think your method works perfectly w/ my 2nd question. I worked it out already. But I still don't know how to solve the 1st question.
     
  7. Dec 1, 2009 #6

    Mark44

    Staff: Mentor

    Then I think you're stuck, unless there is some additional information we haven't seen yet. e-x2 doesn't have a nice neat antiderivative.
     
  8. Dec 1, 2009 #7
    The question says Let X1...Xn be a random sample from a Rayleigh distribution with pdf
    f(x) = (x/a)*e-x2/(2a), for x>0. That's it! And I'm suppose to find E(x).
     
  9. Dec 1, 2009 #8

    Dick

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    Science Advisor
    Homework Helper

  10. Dec 1, 2009 #9
    thank you so much!!
     
  11. Dec 1, 2009 #10
    Hi Dick, I read through the article on wiki, and I'm wondering how the parameter a in my problem affect the integral. I cannot ignore a here since the reason I'm calculating E(x) is becoz I'm trying to find the Method of Moment estimator of a.

    Thank you!
     
  12. Dec 1, 2009 #11

    Dick

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    Set u^2=x^2/(2a). So u=x/sqrt(2a). If you write the integral in terms of u, you should be able to collect all of the a's outside of the integral.
     
  13. Dec 1, 2009 #12
    But it still doesn't solve the integral. it became: Integral 2u^2*e^(-u2)sqrt(2a)du
     
  14. Dec 1, 2009 #13

    Dick

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    Isn't it 2*sqrt(2a)*integral u^2*e^(-u^2)*du? The u integral is just a constant. I thought you were going to use integration by parts like Mark44 suggested to reduce it to the Gaussian integral?
     
    Last edited: Dec 1, 2009
  15. Dec 1, 2009 #14
    Thank you very much!
     
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